(i) Find the gradient of line AB:

The gradient \(m\) of line \(AB\) is given by:

\(m = \frac{3a + 9 - (2a - 1)}{2a + 4 - a} = \frac{a + 10}{a + 4}\)

The gradient of a line perpendicular to \(AB\) is the negative reciprocal:

\(m_{\perp} = -\frac{a + 4}{a + 10}\)

(ii) Find the possible values of \(a\):

The distance \(AB\) is given by:

\(\sqrt{(2a + 4 - a)^2 + (3a + 9 - (2a - 1))^2} = \sqrt{260}\)

Simplify the expression:

\(\sqrt{(a + 4)^2 + (a + 10)^2} = \sqrt{260}\)

Square both sides:

\((a + 4)^2 + (a + 10)^2 = 260\)

Expand and simplify:

\(a^2 + 8a + 16 + a^2 + 20a + 100 = 260\) \(2a^2 + 28a + 116 = 260\) \(2a^2 + 28a - 144 = 0\)

Divide by 2:

\(a^2 + 14a - 72 = 0\)

Factorize:

\((a - 4)(a + 18) = 0\)

Thus, \(a = 4\) or \(a = -18\).